Q.The correct expression representing the relation between thevan't Hoff factor (*i*) and the degree of dissociation (*α*) of a weak electrolyte*MaNb*is

- A) α=ia+ba+b-1
- B) α=ia+ba+b+1
- C) α=i-1a+b-1
- D) α=i-1a+b+1

Q.The correct expression representing the relation between thevan't Hoff factor (*i*) and the degree of dissociation (*α*) of a weak electrolyte*MaNb*is

- A) α=ia+ba+b-1
- B) α=ia+ba+b+1
- C) α=i-1a+b-1
- D) α=i-1a+b+1

Please find the solution to the asked query:

The electrolyte given is M

_{a}N

_{b}

So dissociation of the electrolyte gives:

$GiventhatVantHofffactor=i\phantom{\rule{0ex}{0ex}}Degreeofdissociation=\alpha \phantom{\rule{0ex}{0ex}}Thus\phantom{\rule{0ex}{0ex}}{M}_{a}{N}_{b}\leftrightarrow a{M}^{b+}+b{N}^{a-}\phantom{\rule{0ex}{0ex}}1-\alpha \alpha a\alpha b\phantom{\rule{0ex}{0ex}}Thus\phantom{\rule{0ex}{0ex}}Totalamountofspecies=1+(a+b-1)\alpha \phantom{\rule{0ex}{0ex}}VantHofffactori=\frac{1+(a+b-1)\alpha}{1}\phantom{\rule{0ex}{0ex}}\alpha =\frac{i-1}{(a+b-1)}$

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